Ok percent error is abs(calculated-actual)/actual(100%)
So 1.5/96 *100%
The term "valence electrons" refers to all of the electrons in an atom's outermost shell.
The centre of the atom is where the nucleus is. The nucleus contains protons and neutrons. The electrons travel in a specific circular direction and at a specific distance from the nucleus.
The atom's final shell's electrons take part in chemical reactions and the production of bonds. Both ionic and covalent bonding involve valence electrons. Metals are elements with one, two, or three electrons in their final shell.
These substances become positive ions after losing their electrons. Non-metals are substances with 5, 6, or 7 electrons in the outermost shell. These substances all gain electrons and change into negative ions.
Ionic bonds are those created by the transfer of electrons between metals and non-metals. For instance, ionic bonding allows sodium and chlorine to interact to generate sodium chloride.
To know more about valence electrons here
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Answer:
- <u>2.59 × 10⁻⁷ m = 259 nm</u>
Explanation:
You need to calculate the wavelength of a photon with an energy equal to 463 kJ/mol, which is the energy to break an oxygen-hydrogen atom.
The energy of a photon and its wavelength are related by the Planck - Einstein equation:
Where:
- h = Planck constant (6.626 × 10⁻³⁴ J . s) and
- ν = frequency of the photon.
And:
Where:
- c = speed of light (3.00 × 10⁸ m/s in vacuum)
- λ = wavelength of the photon
Thus, you can derive:
Solve for λ:
Before substituting the values, convert the energy, 463 kJ/ mol, to J/bond
- 463 kJ/ mol × 1,000 J/kJ × 1 mol / 6.022 × 10 ²³ atom × 1 bond / atom
= 7.69×10²³ J / bond
Substitute the values and use the energy of one bond:
- λ = 6.626 × 10⁻³⁴ J . s × 3.00 × 10⁸ m/s / 7.69×10²³ J = 2.59 × 10⁻⁷ m
The wavelength of light is usually shown in nanometers:
- 2.59 × 10⁻⁷ m × 10⁹ nm / m = 259 nm ← answer
Answer:
The balanced equations for those dissociations are:
Ba(OH)₂(aq) → Ba²⁺(aq) + 2OH⁻ (aq)
H₂SO₄ (aq) → 2H⁺(aq) + SO₄⁻²(aq)
Explanation:
As a strong base, the barium hidroxide gives OH⁻ to the solution
As a strong acid, the sulfuric acid gives H⁺ to the solution
Ba(OH)₂, is a strong base so the dissociation is complete.
H₂SO₄ is considerd a strong acid, but only the first deprotonation is strong.
The second proton that is released, has a weak dissociation.
H₂SO₄ (aq) → H⁺(aq) + HSO₄⁻(aq)
HSO₄⁻(aq) ⇄ H⁺ (aq) + SO₄⁻² (aq) Ka
